**Example 13: A problem leading to the case “squares = roots”** [Rosen 1831, 35–36]

*I have divided ten into two portions; I have multiplied the one of the two portions by the other; after this I have multiplied the one of the two by itself, and the product of the multiplication by itself is four times as much as that of one of the portions by the other.*

*Computation: Suppose one of the portions to be thing, and the other ten minus thing: you multiply thing by ten minus thing; it is ten things minus a square.*

*Then multiply it by four, because the instance states "four times as much." The result will be four times the product of one of the parts multiplied by the other. This is forty things minus four squares.*

*After this you multiply thing by thing, that is to say, one of the portions by itself. This is a square, which is equal to forty things minus four squares.*

*Reduce it now by the four squares, and add them to the one square. Then the equation is: forty things are equal to five squares;*

*and one square will be equal to eight roots, that is, sixty-four; *

*the root of this is eight, and this is one of the two portions, namely, that which is to be multiplied by itself. *

*The remainder from the ten is two, and that is the other portion. *

*Thus the question leads you to one of the six cases, namely, that of “squares equal to roots.” *

Continue to Example 14.

Back to Overview of al-Khwārizmī’s More Complicated Problems.